The equation of the circle with the origin as | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For an equilateral triangle,the centroid coincides with the circumcentre.Given the origin $(0, 0)$ is the centre of the circle passing through the

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) For an equilateral triangle,the centroid coincides with the circumcentre.Given the origin $(0, 0)$ is the centre of the circle passing through the vertices,it is the circumcentre of the triangle.The centroid divides the median in the ratio $2:1$.Since the length of the median is $3a$,the distance from the centroid to the vertex (which is the radius $R$ of the circumcircle) is $R = \frac{2}{3} 	imes 3a = 2a$.The equation of the circle with centre $(0, 0)$ and radius $R = 2a$ is $x^2 + y^2 = (2a)^2 = 4a^2$.Since $4a^2$ is not among the given options,the correct answer is $(d)$.

The equation of the circle with the origin as the centre passing through the vertices of an equilateral triangle whose median is of length $3a$ is

Similar Questions

Find the equation of a circle that touches the lines $x = 0$,$y = 0$,and $x = 2c$.

The abscissae of the two points $A$ and $B$ are the roots of the equation $x^2+2ax-b^2=0$ and their ordinates are roots of the equation $y^2+2py-q^2=0$. Then the equation of the circle with $AB$ as diameter is given by

The equation of the circle whose diameter is the line segment joining the points $(-4, 3)$ and $(12, -1)$ is:

If two diameters of a circle of circumference $10 \pi$ lie along the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$,then the equation of the circle is

The equation of the circle touching $x = 0$,$y = 0$ and $x = 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module